CCommons
Search✍️ WriteSign in
⚛️ NEET Physics❓ Question

Potential is zero on the equatorial line of a dipole but field isn't, how can no work be done then?

Siddharth RaoDropper Sample

4 Oct 2026 · 1 min read

Second attempt, going back to basics and trying not to just mug formulas this time. Electrostatic Potential chapter.

For an electric dipole, the potential at a point on the equatorial line is zero (since V=pcos⁡θ4πε0r2V = \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2} and θ=90∘\theta = 90^\circ). But the electric field there is definitely not zero, it's E=p4πε0r3E = \dfrac{p}{4\pi\varepsilon_0 r^3} directed opposite to the dipole moment.

So I have a point where V=0V=0 but E≠0E \ne 0. If there is a field, there is a force on a test charge, so why do we say no work is done in taking a charge from infinity to a point on the equatorial line?

I did write one sentence that potential at a point is the work done in bringing a unit positive charge from infinity, and I understand it for a single charge. But I'm not getting the picture for the dipole equatorial line. Is the equatorial plane an equipotential surface, and if yes, isn't that weird with a field present?

Also, what's the relation between field and potential I should have in my head so I stop making this mistake?

WhatsAppPost on X

6 answers & replies

Sign in to join the conversation.

  • [+]Arjun Nair12thSample· 16d ago

    The key is that field and potential are linked by E⃗=−∇V\vec{E} = -\nabla V, which is the slope of V, not its value. V=0V=0 at a point does not tell you the slope there. Example: halfway between +q+q and −q-q on the line joining them, V is zero but E is not.

    On the equatorial plane V is zero everywhere on the plane (an equipotential). Field lines are perpendicular to equipotentials, so E there points straight along the axis direction, perpendicular to the plane. Moving a charge along the plane means F⃗⋅dl⃗=0\vec{F}\cdot d\vec{l} = 0 at each step, so zero work.

    • [+]Siddharth RaoDropperSample· 15d agoAuthor

      The contour map idea plus slope vs value finally did it for me. I kept confusing the value with the gradient. Thanks all, this thread is going in my notebook.

  • [+]Neel Chatterjee12thSample· 16d ago

    Another way to see the "bring from infinity" bit. Bring the test charge in along the equatorial line from infinity. At every point the force from the dipole is perpendicular to the equatorial line... wait, along the line itself the force has no component along the path? Yes. Force from the field is perpendicular to the path direction, so each small bit of work is F⃗⋅dl⃗=0\vec{F}\cdot d\vec{l}=0. Total work is zero, so V=0V=0. The geometry of the two charges makes this happen.

  • [+]Aman YadavDropperSample· 16d ago

    Simple. V=0V = 0 does not imply E=0E=0. E=0E = 0 does not imply V=0V = 0 either. Inside a charged conducting shell, E is zero but V equals the surface value. Learn both examples, NEET asks it as a statement type question.

  • [+]Rahul DubeyDropperSample· 16d ago

    Siddharth, we have the same habit. I treated "potential" like "force" for two years.

    What worked: potential is an energy-per-charge map. Equipotential lines are like contour lines on a hill map. Field is the steepness. A flat plateau at height zero is possible on a hillside that's steep around it. On the equatorial plane the map is flat along the plane (V=0 everywhere), but steep if you go off the plane toward the positive end (V>0) or negative end (V<0). The force pushes you downhill, which is perpendicular to the plane.

  • [+]Ishita GoyalDropperSample· 15d ago

    Little checklist I keep taped near my desk:

    1. Work done by field =−q ΔV= -q\,\Delta V (so zero if endpoints have same V, regardless of path)
    2. Work by external agent (slow, no KE) =+q ΔV= +q\,\Delta V
    3. Potential at P = work by external agent to bring +1 C from infinity to P
    4. E=−dV/drE = -dV/dr

    If someone gives you a numerical and you see "V same at start and end", the work by the field is zero. Don't compute forces at all, saves time.

More from Siddharth Rao

Keep reading